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    Question

      `sqrt(sqrt(20+sqrt(20+sqrt(20)) ... prop)`   =

    ?
    A 4 Correct Answer Incorrect Answer
    B 2 Correct Answer Incorrect Answer
    C 5 Correct Answer Incorrect Answer
    D 6 Correct Answer Incorrect Answer

    Solution

    let x = `sqrt(sqrt(20+sqrt(20+sqrt(20)) ... prop)` So x = √20+x So x2 = 20+x or x2-x-20 =0 or x = -4,5. x cannot be negative so final answer is 5. Alternate Method : For these type of questions of positive infinite, just brake the number in the product of two consecutive number and answer will be bigger one. 20 = 4 `xx` 5 so answer is 5. If the given number cannot be break in two consecutive numbers then for x=   `sqrt(sqrt(N+sqrt(N+sqrt(N)) ... prop)` , use this shortcut ,  x =  `{1 + sqrt(4N+1)}/2`

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